See Our New JEE Book on Amazon

A point source of heat of power P is placed : Heat Conduction

By

Problem: A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T, the thickness of the shell should not be less than______.  (IIT JEE 1991)

Answer: The answer is $4\pi R^2 KT/P$.

Solution: The heat produced per unit time by the source is $P$. This heat is distributed over inner surface of the spherical shell having area $A=4\pi R^2$. Let $x$ be thickness of the shell. The rate of heat transfer due to conduction is, \begin{align} {\mathrm{d}Q}/{\mathrm{d}t}={K A T}/{x}, \end{align} where $T$ is temperature difference between inner and outer surfaces of the shell. For the temperature $T$ to be constant, the rate of heat incident on the inner surface of the shell should be equal to the rate of heat transfer through conduction i.e., $P={\mathrm{d}Q}/{\mathrm{d}t}$. Use above equation to get, \begin{align} x={KAT}/{P}={4\pi R^2 KT}/{P}.\nonumber \end{align}

More on Heat Conduction

See Our Book

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
JEE Physics Solved Problems in Mechanics